I think it is false right now ;o Take $\{ a_i \}$ to be 2016 $1$s followed by a zero, then 2017 $1$s followed by a zero, then 2018 $1s$, and so on. Clearly this will not be periodic. Perhaps the problem meant $a_{k-1}+ a_{k-2} + ...+ a_{k-2016} < 23$ instead

rmrf wrote: I think it is false right now ;o Take $\{ a_i \}$ to be 2016 $1$s followed by a zero, then 2017 $1$s followed by a zero, then 2018 $1s$, and so on. Clearly this will not be periodic. Perhaps the problem meant $a_{k-1}+ a_{k-2} + ...+ a_{k-2016} < 23$ instead Notice that we have an iff statement (not if). So your construction is not valid since we cannot have $2017$ consecutive $1$'s because after $2016$ of them the following one should be $0$.

This seems too simple... Define a block as 2016 consecutive numbers in the sequence. Given any certain block, a certain sequence must follow. The problem essentially asks to prove that the sequence is periodic after some $a_N$. There are a finite amount of possible blocks, $2^{2016}$. Thus, we will eventually have two identical blocks appear in the sequence - let the starts of each of these blocks be $a_m$ and $a_n$ respectively. Since every next integer in the sequence is defined by only the 2016 subsequent integers, the period will be the length between these two blocks, $n-m$. Thus, $N=m$ and $T=m-n$. We are done.

We also can prove that the period is $2017$ by showing that $$a_k=0 \iff 24-a_{k-2017}>23 \iff a_{k-2017}=0,$$for large enough $k$.

shinichiman wrote: We also can prove that the period is $2017$ by showing that $$a_k=0 \iff 24-a_{k-2017}>23 \iff a_{k-2017}=0,$$for large enough $k$. Can you show a solution please? Thanks!!!

1 year later...

We are going to show $T=2017$ works for $k$ large enough. If there are no such consecutive $(a_k,a_{k+1})=(1,0)$ for some $k>N>4034$ our sequence will be a fixed sequence after a while. So we can take $T=1$. Let's assume $(a_k,a_{k+1})=(1,0)$ for some $k>N$. That means we have; \[ a_k=1 \quad \Longleftrightarrow \quad a_{k-1}+a_{k-2}+\cdots+a_{k-2016}\leq 23 \]\[ a_{k+1}=0 \quad \Longleftrightarrow \quad 1+a_{k-1}+a_{k-2}+\cdots+a_{k-2015}\geq 24 \]which implies $a_{k-2016}=0=a_{k+1}$. Now assume $a_m=a_{m-2017}$ for all $k<m<n$. We are going to show that $a_n=a_{n-2017}$ by induction. \[ a_{n-1}+a_{n-2}+\cdots+a_{n-2016}=a_{n-2018}+a_{n-2019}+\cdots+a_{n-4033} \implies a_n=a_{n-2017} \]